【Leetcode】500. Keyboard Row

题目地址：

https://leetcode.com/problems/keyboard-row/

给定一个字符串数组，返回其中可以由键盘中其中一行打出来的所有字符串。

先把键盘的每一行存入三个哈希表，然后遍历数组。可以用一个数$x$的二进制位来记录每个字符串用了那几行，然后判断一下$x$是否等于$1$、$2$或者$8$即可。代码如下：

import java.util.*;

public class Solution {
    public String[] findWords(String[] words) {
        List<String> ans = new ArrayList<>();
        Set<Character>[] sets = (HashSet<Character>[]) new HashSet[3];
        String[] rows = {"qwertyuiop", "asdfghjkl", "zxcvbnm"};
        for (int i = 0; i < rows.length; i++) {
            sets[i] = new HashSet<>();
            for (int j = 0; j < rows[i].length(); j++) {
                sets[i].add(rows[i].charAt(j));
            }
        }
    
        for (int i = 0; i < words.length; i++) {
            int row = 0;
            for (int j = 0; j < words[i].length(); j++) {
                char ch = Character.toLowerCase(words[i].charAt(j));
                // 如果发现ch在其中某一行，就将那一行的编号的二进制位置为1
                if (sets[0].contains(ch)) {
                    row |= 1;
                }
                if (sets[1].contains(ch)) {
                    row |= 2;
                }
                if (sets[2].contains(ch)) {
                    row |= 8;
                }
            }
            
            if (row == 1 || row == 2 || row == 8) {
                ans.add(words[i]);
            }
        }
        
        String[] res = new String[ans.size()];
        for (int i = 0; i < res.length; i++) {
            res[i] = ans.get(i);
        }
        
        return res;
    }
}

时间复杂度$O(nl)$，空间$O(nl)$。